Modern Physics 5 Question 29
32. The isotopes ${ } _5^{12}$ B having a mass $12.014 u$ undergoes $\beta$-decay to ${ } _6^{12} C .{ } _6^{12} Chas$ an excited state of the nucleus $\left({ } _6^{12} C^{*}\right)$ at 4.041 $MeV$ above its ground state. If ${ } _5^{12} B$ decays to ${ } _5^{12} C$, the maximum kinetic energy of the $\beta$-particle in units of $MeV$ is ( $1 u=931.5 MeV / c^{2}$, where $c$ is the speed of light in vacuum)
(2016 Adv.)
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Solution:
- ${ } _5^{12} B \rightarrow{ } _6^{13} C+{ } _{-1}^{0} e+\bar{\nu}$
Mass of ${ } _6^{12} C=12.000 u$ (by definition of 1 a.m.u.)
$Q$-value of reaction,
$$ \begin{aligned} Q= & \left(M _B-M _C\right) \times c^{2}=(12.014-12.000) \times 931.5 \\ = & 13.041 MeV \\ & 4.041 MeV \text { of energy is taken by }{ } _{65}^{12} C^{*} \\ \Rightarrow & \text { Maximum KE of } \beta \text {-particle is }(13.041-4.041)=9 MeV \end{aligned} $$
